#110. Multiplication against RMB,

#110. Multiplication against RMB,
#110. Multiplication back-to-source memory limit: 256 MiB time limit: 1000 ms standard input and output question type: traditional evaluation method: text comparison uploaded by: Menci submit Record statistics test data question description

This is a template question.

Given positive integers n nn and p pp, evaluate all numbers in 1 percentile n 1 \ sim n1 limit n in the sense of p pp.

Input Format

One row has two positive integers, n nn and p pp.

Output Format

Line n nn, line I ii is a positive integer, indicating the multiplication inverse element of I ii in the sense of p pp.

Sample Input

10 13

Sample output

179108112534

Data range and prompt

1 ≤ n ≤ 3 × 106, n <p <20000528 1 \ leq n \ leq 3 \ times 10 ^ 6, n <p <200005281 ≤ n ≤ 3 × 106, n <p <20000528
P pp is the prime number.

Because p is a prime number, we can easily think of a fast power. However, the last point of the quick power will include TLE 1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 # include <queue> 6 # include <algorithm> 7 # include <cstdlib> 8 # define lli long int 9 using namespace std; 10 const lli MAXN = 10001; 11 void read (lli & n) 12 {13 char c = '+'; lli x = 0, flag = 1; 14 while (c <'0' | c> '9') 15 {c = getchar (); if (c = '-') flag =-1 ;} 16 while (c> = '0' & c <= '9') 17 {x = x * 10 + c-48; c = getchar ();} 18 n = (x * flag); 19} 20 lli n, mod; 21 lli fastpow (lli x, lli n) 22 {23 lli ans = 1; 24 (; n;) 25 {if (n & 1) ans = (ans * x) % mod; x = (x * x) % mod, n = n> 1 ;} 26 return ans; 27} 28 int main () 29 {30 read (n); read (mod); 31 for (lli I = 1; I <= n; I ++) 32 printf ("% lld \ n", fastpow (I, mod-2) % mod); 33 return 0; 34}KSM

Then I read a great blog.

See a recursive formula:

Ans [I] = (mod-mod/I) * ans [mod % I] % mod;

Although I don't know what it means, it should be able to launch reverse yuan very quickly ,,

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<algorithm> 7 #include<cstdlib> 8 #define lli long long int  9 using namespace std;10 const lli MAXN=10000001;11 void read(lli &n)12 {13     char c='+';lli x=0,flag=1;14     while(c<'0'||c>'9')15     {c=getchar();if(c=='-')flag=-1;}16     while(c>='0'&&c<='9')17     {x=x*10+c-48;c=getchar();}18     n=(x*flag);19 }20 lli n,mod;21 int ans[MAXN];22 int main()23 {24     read(n);read(mod);25     ans[1]=1;26     printf("1\n");27     for(int i=2;i<=n;i++)28     {29         ans[i]=(mod-mod/i)*ans[mod%i]%mod;30         printf("%d\n",ans[i]);31     }32     return 0;33 }